Question

raphael graphed the functions $g(x) = x + 2$ and $f(x) = x - 1$. how many units below the $y$-intercept of $g(x)$ is the $y$-intercept of $f(x)$? graph of two lines $g(x)$ (green) and $f(x)$ (blue) on a coordinate plane options: 3 units, -1 units, 2 units, -3 units

Explanation:

Step1: Find y-intercept of \( g(x) \)

For a linear function \( y = mx + b \), the y-intercept is \( b \). For \( g(x)=x + 2 \), when \( x = 0 \), \( g(0)=0 + 2=2 \). So y-intercept of \( g(x) \) is 2.

Step2: Find y-intercept of \( f(x) \)

For \( f(x)=x - 1 \), when \( x = 0 \), \( f(0)=0 - 1=-1 \). So y-intercept of \( f(x) \) is -1.

Step3: Calculate the difference

To find how many units below the y-intercept of \( g(x) \) is the y-intercept of \( f(x) \), we subtract the y-intercept of \( f(x) \) from that of \( g(x) \) (or find the vertical distance). The distance is \( 2-(-1)=3 \)? Wait, no. Wait, the y-intercept of \( g(x) \) is 2 (on the y-axis at (0,2)), and y-intercept of \( f(x) \) is -1 (at (0, -1)). The vertical distance between (0,2) and (0, -1) is \( 2-(-1)=3 \)? Wait, no, wait. Wait, "how many units below" means we find the difference: \( 2-(-1)=3 \)? Wait, no, let's think again. The y-intercept of \( g(x) \) is 2, y-intercept of \( f(x) \) is -1. To find how many units below 2 is -1, we calculate \( 2-(-1)=3 \)? Wait, no, the vertical distance between 2 and -1 on the y-axis: from 2 to -1, we go down 3 units (because 2 - 3 = -1). So the y-intercept of \( f(x) \) is 3 units below that of \( g(x) \).

Answer:

3 units